What is $- 3 x^{4} y^{2} \cdot - x^{0} y^{- 4} \cdot 2 x$ $-3x^4y^2 * -x^0y^-4 * 2x$ ?

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What is $- 3 x^{4} y^{2} \cdot - x^{0} y^{- 4} \cdot 2 x$ $-3x^4y^2 * -x^0y^-4 * 2x$ ?

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Answer 1 · 2017-01-25T21:53:20

$6 x^{5} y^{- 2}$ $6x^5y^-2$

Explanation:

Using the property of exponents, that $a^{b} \cdot a^{c} = a^{b + c}$ $a^b * a^c = a^(b+c)$ :

$- 3 x^{2} y^{2} \cdot - x^{0} y^{- 4} \cdot 2 x = - (- 3) \cdot 2 \cdot x^{4 + 1 - 0} \cdot y^{2 - 4}$ $-3x^2y^2 * -x^0y^-4 * 2x = -(-3) * 2 * x^(4 + 1 - 0) * y^(2-4)$

$= 6 x^{5} y^{- 2}$ $= 6x^5y^-2$ or $\frac{6 x^{5}}{y^{2}}$ $(6x^5)/y^2$ . Note that the minus sign in front of $x^{0}$ $x^0$ does matter, even if it's attached to $x^{0}$ $x^0$ , which is equal to $1$ $1$ anyway.

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What is $- 3 x^{4} y^{2} \cdot - x^{0} y^{- 4} \cdot 2 x$ $-3x^4y^2 * -x^0y^-4 * 2x$ ?

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